Example of $Y = A \cup B \cup C$ such that reduced homology groups are trivial for all intersections but not for $Y$ I'm trying to solve the following problem:
Give an example of $Y = A \cup B\cup C$ such that $A,B,C$ are open subsets of $Y$ and the reduced homology groups of $A,B,C, A\cap B, A\cap C, B \cap C, A \cap B \cap C$ are all trivial, the sets $A \cap B, A \cap C , B \cap C$ are non-empty but $H_1 (Y)$ is not trivial. Show that $H_n (Y)$ has to be trivial for $n \geqslant 2$.
It feels like the second part should follow from some exact sequence, but I've tried Mayer-Vietoris to no avail (it's likely i'm missing something though?). As for the example I've tried different partitions of $S^1$ or the torus but I couldn't get it to work. I don't think I have a good enough understanding of what trivial homology groups imply geometrically to come up with an example, other than randomly stumbling onto one. Any help's appreciated.
 A: For the first part, take $Y$ to be the circle $S^1$ and $A$, $B$ and $C$ to be three connected open segments that cover the circle without any unnecessary overlaps, e.g., $(0^{\circ}, 121^{\circ})$, $(120^{\circ}, 241^{\circ})$ and $(240^{\circ}, 1^{\circ})$. Then $A$, $B$, $C$, $A \cap B$, $A \cap C$ and $B \cap C$ are all contractible and $A \cap B \cap C$ is empty, so these spaces all have trivial reduced homology groups. However the circle $Y = S^1 = A \cup B \cup C$ has $\widetilde{H}_1(Y) = \Bbb{Z}$. This non-trivial $\widetilde{H}_1$ sneaks in because when we look at the Mayer-Vietoris sequence for $Y$ viewed as the union of $A \cup B$ and $C$, we find that $\widetilde{H}_1(Y) \cong \widetilde{H}_0((A \cup B) \cap C)$ and the latter group is $\Bbb{Z}$ (because $(A \cup B) \cap C$ has two connected components).
For the second part, use the Mayer-Vietoris sequence first for the union $A \cup B$ and then for the union $Y = A \cup B \cup C$. See Reduced homology groups of a space which is the union of finitely many open subsets for a generalisation.
